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Question

Sum of the first n terms of the following series 13+33+53+73+ is


A

n22n2-1

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B

n3n-1

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C

n3+8n+4

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D

2n4+3n2

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Solution

The correct option is A

n22n2-1


Explanation for the correct option :

Step 1 : Reforming the given series

Let S be the sum of the first n terms of the given series 13+33+53+73+ i.e. S=13+33+53+73++2n-13. We can write S as

S=r=1n2r-13=r=1n8r3-12r2+6r-1=r=1n8r3-r=1n12r2+r=1n6r-r=1n1

Step 2 : Calculating the above summations

Formula to be used : We know that the sum of the first nnatural numbers is nn+12, the sum of the squares of the first nnatural numbers is nn+12n+16 and the sum of the cubes of the first nnatural numbers is nn+122.

So,

r=1n6r=6×nn+12=3nn+1

r=1n12r2=12×nn+12n+16=2nn+12n+1

r=1n8r3=8×nn+122=2n2n+12

Also,

r=1n1=n

Step 3 : Finding the value of S

From Step-1, using the values obtained in Step-2, we can get the value of Sas follows :

S=r=1n8r3-r=1n12r2+r=1n6r-r=1n1=2n2n+12-2nn+12n+1+3nn+1-n=2nn+1n2+n-2n-1+n3n+3-1=2nn+1n2-n-1+n3n+2=2nn+1n2-n-2nn+1+n3n+2=2n2n2-1-2n2-2n+3n2+2n=2n4-2n2+n2=2n4-n2=n22n2-1

Hence, option (A) is the correct answer.


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